Transition metals

The transition metals sit in the middle block of the periodic table, between groups 2 and 3. Compared with group 1 metals they are hard and dense, have high melting points and react slowly or not at all with water. They form ions with different charges, coloured compounds and complex ions, and they make good catalysts. At A-level a transition metal is an element that forms at least one stable ion with an incomplete d sub-shell.

Key facts

  • Typical properties: high melting points and densities, hard and strong, much less reactive than group 1; ions with different charges (Fe²⁺ and Fe³⁺); coloured compounds; good catalysts.
  • The 4s sub-shell fills before 3d, but ions lose their 4s electrons first: Fe is [Ar] 3d⁶ 4s² and Fe³⁺ is [Ar] 3d⁵. Chromium ([Ar] 3d⁵ 4s¹) and copper ([Ar] 3d¹⁰ 4s¹) are the exceptions.
  • Scandium and zinc are in the d block but are not transition metals: Sc³⁺ has no d electrons and Zn²⁺ has a full 3d sub-shell.
  • A complex is a central metal ion joined to ligands by co-ordinate bonds. A ligand donates a lone pair of electrons: water, ammonia, chloride and cyanide are common ligands. The co-ordination number is the number of co-ordinate bonds.
  • Shapes: six small ligands give an octahedral complex (90°); four large ligands such as Cl⁻ give a tetrahedral one (109.5°); platinum(II) complexes such as cisplatin are square planar; silver(I) forms linear [Ag(NH₃)₂]⁺.
  • Colour: the ligands split the d orbitals. An electron absorbs visible light of energy ΔE = hν to move up, and the light that is left is the colour you see. Changing the oxidation state, the ligands or the co-ordination number changes the colour.
  • Catalysts: a heterogeneous catalyst is in a different phase from the reactants (iron in the Haber process, V₂O₅ in the Contact process); a homogeneous catalyst is in the same phase and works by changing its oxidation state (Fe²⁺ in the reaction of I⁻ with S₂O₈²⁻).

Worked examples

Example 1

Work out the oxidation state of the metal in the complex [CoCl₄]²⁻ and in the dichromate ion, Cr₂O₇²⁻.

  1. Each chloride ligand has a charge of −1, so the four ligands add up to −4. 4 × (−1) = −4
  2. The complex has an overall charge of −2, so cobalt is +2. Co = −2 − (−4) = +2
  3. In Cr₂O₇²⁻ each oxygen is −2. 7 × (−2) = −14
  4. The two chromium atoms add up to the rest; each one is +6. 2 × Cr = −2 − (−14) = +12, so Cr = +6

Answer: +2 in [CoCl₄]²⁻; +6 in Cr₂O₇²⁻

Example 2

Write the electron configurations of an iron atom and of the Fe³⁺ ion. Explain why iron is a transition metal.

  1. Iron has 26 electrons: 18 as in argon, then 8 more. 26 − 18 = 8
  2. 4s fills before 3d, so two go into 4s and six into 3d: [Ar] 3d⁶ 4s².
  3. Fe³⁺ has lost 3 electrons. The two 4s electrons go first, then one from 3d: [Ar] 3d⁵. 8 − 3 = 5
  4. The 3d sub-shell of Fe³⁺ holds 5 of a possible 10 electrons: incomplete, so iron is a transition metal.

Answer: Fe: [Ar] 3d⁶ 4s²; Fe³⁺: [Ar] 3d⁵. Fe³⁺ has an incomplete 3d sub-shell.

Example 3

Copper(II) sulfate solution contains blue [Cu(H₂O)₆]²⁺ ions. Concentrated hydrochloric acid is added. What do you see, and why does the shape of the complex change?

  1. The chloride ions replace the water ligands. [Cu(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CuCl₄]²⁻ + 6H₂O
  2. Chloride ions are much larger than water molecules, so only four fit round the copper ion: the co-ordination number falls from 6 to 4 and the shape changes from octahedral to tetrahedral.
  3. Different ligands and a different co-ordination number split the d orbitals by a different amount, so a different colour of light is absorbed: [CuCl₄]²⁻ is yellow, and mixed with some blue complex the solution looks green.

Answer: The solution turns yellow-green as [CuCl₄]²⁻ forms; the octahedral complex becomes tetrahedral because Cl⁻ ligands are larger than water molecules.